javascript - 如何有效检查连续数字列表是否缺少任何元素

我有这个数组

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];

我试图找到一种算法，该算法可以告诉我缺少哪些s。如您所见，该列表由连续的s(s1，s2等)组成。

首先，我采用了以下解决方案:

    var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (var i=1;i<arr.length;i++){
    var thisI = parseInt(arr[i].toLowerCase().split("s")[1]);
    var prevI = parseInt(arr[i-1].toLowerCase().split("s")[1]);
    if (thisI != prevI+1)
      console.log(`Seems like ${prevI+1} is missing. thisI is ${thisI} and prevI is ${prevI}`)
}

但是，此方法因缺少多个连续的数字(s15，s16)而失败。因此，我添加了一个有效的while循环。

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (var i=1;i<arr.length;i++){
  var thisI = parseInt(arr[i].toLowerCase().split("s")[1]);
  var prevI = parseInt(arr[i-1].toLowerCase().split("s")[1]);
  if (thisI != prevI+1) {
    while(thisI-1 !== prevI++){
       console.log(`Seems like ${prevI} is missing. thisI is ${thisI} and prevI is ${prevI}`)
    }
   }
}

但是，我觉得事情太复杂了。
我想到了创建理想的数组:

var idealArray = [];
for (var i =0; i<200;i++) {
  idealArray.push(i)
}

然后，在检查时，篡改我的数组(arr)，以便循环检查长度相同的两个数组。即，使用以下解决方案:

var idealArray = [];
for (var i =0; i<200;i++) {
  idealArray.push(i)
}
var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (let i = 0; i<idealArray.length;i++){
  if (parseInt(arr[i].toLowerCase().split("s")[1]) != idealArray[i]) {
    console.log(`Seems like ${idealArray[i]}is missing`);
    arr.splice(i,0,"dummyel")
  }
}

但是，我再次感到创建第二个数组也不是很有效(想想一个大列表，我会浪费不必要的空间)。

那么...我该如何在JavaScript中高效地执行此任务？ (有效地表示时间复杂度和空间复杂度都尽可能接近O(1)。)

最佳答案

既然您知道您期望有一个顺序数组，那么我不知道为什么它需要比通过数字arr[0]到arr[end]的循环来更复杂，同时还要保持一个计数器来知道您在数组中的位置。这将以O(n)运行，但我认为您无法对此进行改进-在最坏的情况下，您至少需要查看每个元素一次。

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];

let first = parseInt(arr[0].substring(1))
let last =  parseInt(arr[arr.length-1].substring(1))
let count = 0
for (let i = first; i< last; i++) {
   if (parseInt(arr[count].substring(1)) == i) {count++; continue}
   else console.log(`seem to be missing ${'s'+i.toString().padStart(2,'0')} between: ${arr[count-1]} and ${arr[count]}` )
}

编辑:

在仔细考虑了以下注释之后，我采取了一种递归方法，将数组拆分并检查每一半。主要是作为实验，而不是实际的解决方案。实际上，在大多数情况下，这样做的重复次数少于n，但我找不到实际情况下速度更快的情况。此外，我只是推送了索引，以显示差距在哪里，从而使结构更易于查看和测试。就像您将看到的那样，因为它是递归的，所以结果并不整齐。

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];

let missingGaps = []

function missing(arr, low, high) {
  if (high <= low) return

  let l = parseInt(arr[low].substring(1))
  let h = parseInt(arr[high].substring(1))

  if (h - l == high - low) return
  if (high - low === 1) {
    missingGaps.push([low, high])
    return
  } else {
    let mid = ((high - low) >> 1) + low

    missing(arr, low, mid)

    // need to check case where split might contain gap
    let m = parseInt(arr[mid].substring(1))
    let m1 = parseInt(arr[mid + 1].substring(1))
    if (m1 - m !== 1) missingGaps.push([mid, mid + 1])

    missing(arr, mid + 1, high)
  }
}

missing(arr, 0, arr.length-1)
missingGaps.forEach(g => console.log(`missing between indices ${arr[g[0]]} and ${arr[g[1]]}`))

也许另一个答案或评论将有所改进，使其速度更快。